Find the coordinates of the point on the curve y=x³-3x²+6x+2 at which the gradient is 3

Answer 1

To find the coordinates of the point on the curve where the gradient is 3, we need to find the derivative of the curve and set it equal to 3.

First, we differentiate the equation of the curve y = x³ - 3x² + 6x + 2 with respect to x:

dy/dx = 3x² - 6x + 6

Next, we set the derivative equal to 3:

3x² - 6x + 6 = 3

Simplifying this equation, we get:

3x² - 6x + 3 = 0

Divide through by 3:

x² - 2x + 1 = 0

This equation can be factored as:

(x - 1)² = 0

Therefore, the value of x is 1.

Now, to find the y-coordinate of the point on the curve, substitute x = 1 into the original equation:

y = 1³ - 3(1)² + 6(1) + 2
y = 1 - 3 + 6 + 2
y = 6

Therefore, the coordinates of the point on the curve where the gradient is 3 are (1, 6).